Pair of shorts costs $12. Shirt costs $48.
Difference between the two is $36.
Answer: 15/32
Step-by-step explanation:
The data that we have is:
5/8 of the vegetable sold where green.
of theses 5/8, 3/4 were string beans.
Then the 3/4 is applied to the 5/8 of green vegetables.
Then the fraction (of the total number of vegetables) that were string beans was:
F = (3/4)*(5/8) = (3*5)/(4*8) = 15/32
Answer:
Step-by-step explanation:
- Option A
![[x+5=20]](https://tex.z-dn.net/?f=%5Bx%2B5%3D20%5D)
tells us that: When we add 5 to a variable x, we get 20. As it has a unique value for x and is completely equal to it(i.e. 15), It is an equality.
- Option B
![[x=5]](https://tex.z-dn.net/?f=%5Bx%3D5%5D)
tells us that: A variable x equals to 5. Hence, as x is unique for 5 and is wholly equal to it, it's an equality too. - Option C
tells us that: A variable x isn't 5 but lesser than it. As we cannot equate it to 5, nor we are given the nature of the variable x, it is an Inequality. - Option D
![[x+5]](https://tex.z-dn.net/?f=%5Bx%2B5%5D)
is an expression; It can't be called an equation or an inequality unless we relate it with another expression.
Check the picture below.
since the diameter of the cone is 6", then its radius is half that or 3", so getting the volume of only the cone, not the top.
1)
![\bf \textit{volume of a cone}\\\\ V=\cfrac{\pi r^2 h}{3}~~ \begin{cases} r=radius\\ h=height\\[-0.5em] \hrulefill\\ r=3\\ h=4 \end{cases}\implies V=\cfrac{\pi (3)^2(4)}{3}\implies V=12\pi \implies V\approx 37.7](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bvolume%20of%20a%20cone%7D%5C%5C%5C%5C%20V%3D%5Ccfrac%7B%5Cpi%20r%5E2%20h%7D%7B3%7D~~%20%5Cbegin%7Bcases%7D%20r%3Dradius%5C%5C%20h%3Dheight%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20r%3D3%5C%5C%20h%3D4%20%5Cend%7Bcases%7D%5Cimplies%20V%3D%5Ccfrac%7B%5Cpi%20%283%29%5E2%284%29%7D%7B3%7D%5Cimplies%20V%3D12%5Cpi%20%5Cimplies%20V%5Capprox%2037.7)
2)
now, the top of it, as you notice in the picture, is a semicircle, whose radius is the same as the cone's, 3.
![\bf \textit{volume of a sphere}\\\\ V=\cfrac{4\pi r^3}{3}~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=3 \end{cases}\implies V=\cfrac{4\pi (3)^3}{3}\implies V=36\pi \\\\\\ \stackrel{\textit{half of that for a semisphere}}{V=18\pi }\implies V\approx 56.55](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bvolume%20of%20a%20sphere%7D%5C%5C%5C%5C%20V%3D%5Ccfrac%7B4%5Cpi%20r%5E3%7D%7B3%7D~~%20%5Cbegin%7Bcases%7D%20r%3Dradius%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20r%3D3%20%5Cend%7Bcases%7D%5Cimplies%20V%3D%5Ccfrac%7B4%5Cpi%20%283%29%5E3%7D%7B3%7D%5Cimplies%20V%3D36%5Cpi%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bhalf%20of%20that%20for%20a%20semisphere%7D%7D%7BV%3D18%5Cpi%20%7D%5Cimplies%20V%5Capprox%2056.55)
3)
well, you'll be serving the cone and top combined, 12π + 18π = 30π or about 94.25 in³.
4)
pretty much the same thing, we get the volume of the cone and its top, add them up.
N = no solution
-4 ( 6 - 2n ) = 30 + 8n
-24 + 8n = 30 + 8n
Subtract 8n from both sides
-24 = 30
This is invalid therefore no solution
